Question

(Pfvr)DETERMINE os valores de xe y.​

Answer 1

Resposta:

X=11

Y-3$\sqrt{21}$

Explicação passo-a-passo:

$a^{2}$ = $b^{2}$ + $c^{2}$

$(x)^{2} = (2\sqrt{10}) ^{2} + (9)^{2}$

$x^{2} = 40 + 81$

$x^{2} = 121\\x = {+}{-} \sqrt{121} \\x = + - 11$

$a^{2}$ = $b^{2}$ + $c^{2}$

$(y)^{2} = (2\sqrt{10} )^{2} + (13)^{2} \\y^{2} = 40 + 169\\y^{2} = 209\\y = +-\sqrt{209}$

Fatoração

209 I 3

69  I 3

23  I 3

    7 I 7

    1 I

$\sqrt{209} = \sqrt{3^{2} }.3.7 = \\ y = 3\sqrt{3.7} \\y = 3\sqrt{21}$